Problem: Simplify the following expression: $r = \dfrac{10y^2 + 140y + 480}{y + 8} $
First factor the polynomial in the numerator. We notice that all the terms in the numerator have a common factor of $10$ , so we can rewrite the expression: $ r =\dfrac{10(y^2 + 14y + 48)}{y + 8} $ Then we factor the remaining polynomial: $y^2 + {14}y + {48} $ ${8} + {6} = {14}$ ${8} \times {6} = {48}$ $ (y + {8}) (y + {6}) $ This gives us a factored expression: $\dfrac{10(y + {8}) (y + {6})}{y + 8}$ We can divide the numerator and denominator by $(y - 8)$ on condition that $y \neq -8$ Therefore $r = 10(y + 6); y \neq -8$